From: fatblokeonbike [mailto:[EMAIL PROTECTED]
> $Query="SELECT properties.area, images.image_filename FROM > properties, > images WHERE properties.reference_number=images.reference_number AND > properties.area=$id"; > > but it doesn't work - I get the usual "...not a valid MySQL > result resource" > > I've played around with it, but I confess myself beat. I expect the > answer's terribly simple - but then, as everyone keeps > telling me, so am I. > > If you can help, thanks in advance. If $id is a string and not an int, then you need to single- or double-quote it. Try this: $Query="SELECT properties.area, images.image_filename FROM properties, images WHERE properties.reference_number=images.reference_number AND properties.area='$id'"; HTH! -- Mike Johnson Web Developer/Systems Asst. Smarter Living, Inc. phone (617) 497-2500 x226 -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe: http://lists.mysql.com/[EMAIL PROTECTED]
